4: Vapour Pressure

A non-volatile solute contributes essentially no vapour pressure of its own. Its presence lowers the solvent mole fraction and the escaping tendency of solvent molecules from the liquid surface. By Raoult’s law, p_solution = x_solvent p°_solvent, so the equilibrium vapour pressure falls. The change is not caused by automatic heating or by a zero molar mass.

For a solution containing a non-volatile solute, Raoult’s law gives relative lowering of vapour pressure as (p° − p)/p° = xsolute. Thus, when the solute mole fraction increases, the solvent mole fraction decreases and the vapour pressure falls more. Colour, vessel width and gentle stirring do not determine this colligative quantity.

For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as \((p^0-p)/p^0=x_{\text{solute}}\). The solute does not contribute vapour, so its mole fraction directly measures the fractional reduction in solvent vapour pressure. With \(x_{\text{solute}}=0.05\), the relative lowering is 0.05, or 5%. It is not the solvent fraction 0.95 or the percentage written as 5.0.

For an ideal solution, the partial vapour pressure of each volatile component equals its mole fraction multiplied by its pure-component vapour pressure: pi = xi pi°. In the special case of a non-volatile solute, only the solvent contributes appreciably, so psolution = xsolvent p°solvent. Colour, container height and an always-zero pressure are unrelated claims.

For a non-volatile solute, Raoult’s law is p = xsolvent p°solvent. Substituting xsolvent = 0.8 and p°solvent = 100 gives p = 0.8 × 100 = 80. The pressure must be lower than the pure-solvent value because adding a non-volatile solute reduces the solvent mole fraction; therefore 100 cannot be the solution pressure here.

Relative lowering of vapour pressure is the decrease divided by the vapour pressure of the pure solvent: \((p_1^0-p)/p_1^0\). The decrease is \(120-108=12\). Dividing by the original pressure gives \(12/120=0.10\). Therefore option A is correct. The value 12 is only the absolute decrease, 0.90 is the remaining pressure fraction, and 0.12 uses an incorrect denominator.

For a non-volatile solute, only the solvent contributes significantly to the vapour pressure, and Raoult’s law gives p = xsolvent p°solvent. Thus p = 0.75 × 200 = 150. The result is lower than 200, as expected because the solvent mole fraction is less than one. Dividing by 0.75 would incorrectly reverse the relation.

For a non-volatile solute, Raoult’s law leads to (p° − p)/p° = xsolute. The left side is the relative lowering of vapour pressure. Since it is given as 0.2, the solute mole fraction is also 0.2. The solvent mole fraction would then be 1 − 0.2 = 0.8; that is why 0.8 is a plausible but incorrect distractor.

For a solution containing a non-volatile solute, the solvent alone contributes significantly to the vapour phase. Raoult's law relates the vapour pressure of the solvent in the solution to its mole fraction, so it also gives the relative lowering of vapour pressure. That relation can then be combined with the solute mass to determine molar mass. The other laws belong to mechanics, electricity, and elasticity.

For a solution containing a non-volatile solute, Raoult’s law states that the relative lowering of vapour pressure, (p° − p)/p°, equals the mole fraction of the solute, xB, for a sufficiently dilute solution. Therefore, measuring the vapour-pressure lowering gives the solute mole fraction, which can be combined with the known masses to calculate its molar mass. Option A is correct.

The lowering of vapour pressure method assumes that the vapour above the solution is contributed essentially by the solvent. A non-volatile solute does not itself enter the vapour phase, so the change in solvent vapour pressure can be related to the solute mole fraction. A volatile solute would invalidate this simple treatment. Therefore option A is correct.

For a solution containing a non-volatile solute, Raoult’s law gives relative lowering of vapour pressure as (p° − p)/p° = xsolute. The stated value is 0.02, so the dilute-solution approximation gives xsolute ≈ 0.02. The value 0.98 is approximately the solvent mole fraction, while 2.0 is impossible for a mole fraction. Hence A is correct.

For the same solvent, temperature and nominal concentration, a colligative effect is multiplied by the van't Hoff factor i. If the measured relative lowering is half the normal expected value, the effective particle count is half the normal count, so i ≈ 0.5. Such a value suggests strong association, especially near complete dimerisation.

For a fixed amount of solvent and solute, the ratio of the observed colligative effect to the normal expected effect is i. If the effect is 30% greater, its ratio is 100% + 30% = 130% = 1.30. Therefore i = 1.30. This indicates an increased effective particle number, as would occur when a solute partially dissociates.

A colligative effect that is 40% higher than the normal value is 100% + 40% = 140% of that value, or 1.40 times it. Since the van’t Hoff factor is the ratio of the actual effect to the normal effect under the same conditions, i = 1.40. It is not 0.40, which represents only the increase.